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\chapter*{Remerciement}

Tout d'abord, je voudrais exprimer mon gratitude à tous ceux qui beaucoup m'aider pendant le stage et a rediger le rapport

Je remercie particulièrement à M.Nguyen Anh Ky, mon directeur de stage pour sa disponibilité et son enthousiasme et M.Phi Quang Van. 
Merci de m'avoir beacoup aidé, m'avoir expliqué et avoir repondu à tous mes questions.

Je remercie a Phan Duc Dung qui m'a beaucoup aide pendant le stage et aussi son enthosiasme pour corriger mon rapport. 
Merci pour ses aides ses explications, ses discussion et aussi sa patience.

Je remercie les amis dans mon programmes de Master qui m'a aider beaucoup de choses dans ce processus d'apprendre le Master  

\chapter*{Introduction}

Les paramètres $S,T,U$ ( ou les paramètre de Peskin-Takeuchi) ce sont les trois quantités mesurables pour chercher les contributions de nouvelle physique. 
Dans les modèles de la physiques de particules, les particules de la nouvelle physique sont très lourds.
Par conséquence, ils ne peuvent qu’apparaitre dans la correction boucle.
On observe ces paramètres quand on calcule les corrections obliques des quantités physiques, c’est-à-dire, les énergies propres de particules jauges.

Pour vérifier un modèle quelconque de la théorie des particules.
Un des méthodes le plus populaires est calculés les $S,T,U$ et compare avec l’expérimentale.
On peut prévoir l’existance de la nouvelle physique.
	
Dans ce rapport, donc le temps est seulement 3 mois, je présente seulement la fa\c{c}on de mon calcul et calcule dans le cas qui est le plus facile.
Ce rapport consiste 3 chapitres, une conclusion et un annex. Dans le chapitre 1, je représente la théorie de jauge et le Modèle Standard.
Le chapitre 2, c'est la calculation de la correction une boucle. Et dans le chapitre 3, c'est la calculation les $S,T,U$ et leurs interprétations

    	
\title{Calculer les S,T,U parametre et l'interpretation desquels par la nouvelle physique }

\chapter{Introduction to the Standard Model}
\section{Idea of the gauge invariance}

To understand the basic structure of the Standard Model (SM), we should learn the gauge theory. In this part, 
I will represent a brief review on the idea of the 
gauge invariance. Let's first consider the simplest case, QED, which satisfies a $U(1)$ gauge symmetry. 
We start with the Lagrangian of the electron field $\psi(x)$ with mass $m$, \cite{VSOP15}
\begin{equation}
\label{lagrangian 1}
\mathcal{L}=\bar{\psi}(i{\gamma}_{\alpha}\partial_{\alpha}-m)\psi,
\end{equation}
where $\partial_{\alpha}\equiv\dfrac{\partial}{\partial_\alpha}$. Observe that under the transformation 
$\psi(x)\rightarrow\psi^{\prime}=e^{i{\Lambda}}\psi(x)$ where $\Lambda$ is a real constant, the Lagrangian 
is invariant: $L(\psi)=L(\psi^{\prime})$. Different transformations of group $U(1)$ commute. 
Such groups are called "abelian" and since $\Lambda$ is a constant, this group $U(1)$ is also called "global".

Now suppose that the group is still $U(1)$, but "local" ($\Lambda\equiv\Lambda(x)$). Then under the local 
transformation $\psi^{\prime}=e^{i\Lambda(x)}\psi(x) \equiv U(x)\psi(x)$, the derivative $\partial_{\alpha}\psi(x)$ 
transforms as 
\begin{equation}
\label{transformation}
\begin{split}
\partial_{\alpha}\psi(x) &= \partial_{\alpha}U^{-1}(x)\psi^{\prime}(x)\\
&=U^{-1}(x)U(x)\partial_{\alpha}U^{-1}(x)\psi^{\prime}(x)\\
&=U^{-1}(x)(1+i{\Lambda})\partial_{\alpha}(1-i{\Lambda}\psi^{\prime}(x))\\
&=U^{-1}(\partial_{\alpha}-i{\partial}_{\alpha}\Lambda)\psi^{\prime}.
\end{split}
\end{equation}
From \eqref{transformation} (because ${\Lambda}\neq constant$) we have $\psi(x)=U^{-1}(x)\psi^{\prime}(x)$ 
but $\partial_{\alpha}\psi(x)\neq U^{-1}(x)\partial_{\alpha}\psi^{\prime}(x)$ it means the field and its 
derivative do not transform in the same way under a local transformation. For the global case, if we 
recall, they did transformation in the same way, and the Lagrangian remained invariant. But now for 
the local case, $\mathcal{L}(\psi)\neq\mathcal{L}(\psi^{\prime})$, i.e Lagrangian doesn't invariant.

As the next step, we rewrite the Lagrangian in equation \eqref{lagrangian 1} with $D_{\alpha}\equiv 
\partial_{\alpha}-ieA_{\alpha}(x)$ replacing $\partial_{\alpha}$, where $e$ is a coupling constant, and 
$D_{\alpha}$ is called the covariant derivative. We now observe the following:
\begin{equation}
\begin{split}
[\partial_{\alpha}-ieA_{\alpha}(x)]\psi(x)& = U^{-1}U(x)[\partial_{\alpha}-ieA_{\alpha}(x)]U^{-1}(x)\psi^{\prime}(x)\\
& =U^{-1}(x)[U(x)\partial_{\alpha}U^{-1}(x)-ieU(x)A_{\alpha}(x)U^{-1}(x)]\psi^{\prime}(x)\\
& =U^{-1}(x)[\partial_{\alpha}-i\partial_{\alpha}\Lambda(x)-ieA_{\alpha}(x)]\psi^{\prime(x)}\\
& =U^{-1}(x)[\partial_{\alpha}-ieA^{\prime}_{\alpha}(x)]\psi^{\prime}(x)
\end{split},
\end{equation}
where $A^{\prime}_{\alpha}\equiv A_{\alpha}(x)+\dfrac{1}{e}\partial_{\alpha}\Lambda(x)$. We see that the covariant derivates transforms like the field itself: $D_{\alpha}\psi(x)=U^{-1}(x)D^{\prime}_{\alpha}\psi^{\prime}(x)$ where 
$D'_{\alpha}\equiv\partial_{\alpha}-ie A'_{\alpha}(x)$. This implies that $\mathcal{L}$ of Eq. \eqref{lagrangian 1} 
after replacing $\partial_{\alpha}$ by $D_{\alpha}$, is invariant under the gauge transformation.

Now we concentrate on the non-abelian, i.e., noncommutative, group $SU(2)$. Consider a fermion field $\psi(x)$ 
which transforms as a doublet under $SU(2)$:
 $\psi(x)=\begin{pmatrix}
\psi^1(x)\\
\psi^2(x)
\end{pmatrix}$. Let's follow its local $SU(2)$ transformation:$\psi\rightarrow\psi'=e^{\frac{i}{2}\tau_a\Lambda_a(x)}\psi(x)$, and 
$\bar{\psi}\rightarrow\bar{\psi'}e^{\frac{-i}{2}\tau_a\Lambda_a(x)}$, where $\tau_a(a=1,2,3)$ are the Pauli matrices which satisfy 
$[\tau_a,\tau_b]=2i{\epsilon}_{abc}\tau_c$. It is easy to check that $\partial_{\alpha}\psi(x)=U^{-1}(x)[\partial_{\alpha}-
\frac{i}{2}\tau_{\alpha}\partial_{\alpha}\Lambda_a(x)]\psi'(x)$, we have $\psi(x)$ and $\partial_{\alpha}\psi(x)$ do not transform 
identically, and hence the Lagrangianis is not invariant under $SU(2)$ transformation \cite{VSOP15}. If replacing the derivate $\partial_{\alpha}$ 
by covariant derivative $D_{\alpha}\psi(x)=[\partial_{\alpha}-\frac{ig}{2}\tau_{\alpha}A^a_{\alpha}(x)]\psi(x)$, where $g$ is the coupling 
constant(like the symbol $e$ used for $U(1)$), with $A^a_{\alpha}(x)$ transforming in an appropriate way (see \eqref{gf}), we can obtain 
a Lagrangian invariant under a local $SU(2)$ transformation. 

Note that in the literature the formulas of covariant derivatives 
$D_{\alpha}\psi(x)=[\partial_{\alpha}+\frac{ig}{2}\tau_aA_{\alpha}^a(x)]\psi(x)$ 
with the gauge group $SU(2)$ and $ D_{\alpha}= \partial_{\alpha}+ieA_{\alpha}(x) $ 
with the gauge group $U(1)$ are also used with $i$ replaced by $-i$ in the gauge 
transformation. \cite{Morii2004} 


\section{Spontaneous Symmetry Breaking and Higgs mechanism}
\subsection{Spontaneous Symmetry Breaking}

Let $U$ be an element of the symmetry group which leaves Hamiltonian $H$ (or Lagrangian $\mathcal{L}$) invariant, \cite{Cheng-Li84} 
\begin{equation}
\label{Hamiltonian invariant}
UHU^+=H,
\end{equation}
and it connects states that form an irreducible representation (basis) of the group,
\begin{equation}
\label{eq1.2}
U|a\rangle=|b\rangle .
\end{equation}
That means 
\begin{equation}
\label{eq1.3}
E_b=\langle{b}|H|b\rangle=\langle{a}|U^+HU|a\rangle=E_a.
\end{equation}
Thus the symmetry of Hamiltonian $H$ is manifest in the degeneracies of the energy eigenstates corresponding to irreducible 
representations of the symmetry group. However, since all states are created from vacuum, we have from \eqref{eq1.2}
\begin{equation}
|a\rangle=\psi_a|0\rangle ;~ |b\rangle=\psi_b|b\rangle,
\end{equation}
and
 \begin{equation}
 U^+\psi_aU=\psi_b.
 \end{equation}
Equation \eqref{eq1.2} follows only if
\begin{equation}
\label{invariant du vide}
U|0\rangle=|0\rangle .
\end{equation}
Because groups of symmetry in particle physics are Lie groups, we have
 $U|0\rangle=|0\rangle\ \Leftrightarrow Q_\alpha|0\rangle=|0\rangle$, with $Q_\alpha$ are generators of this transformation. 
When the equation \eqref{invariant du vide} is false, equation \eqref{eq1.3} is also false.  
We say  this symmetry is "hidden" or spontaneous broken. One important consequence of spontaneous breaking is the scalar 
particles with $m=0$ called Goldstone bosons, number of Goldstone bosons = number of generators which do not leave vacuum invariant. 
By Higgs mechanism these Goldstones disappear, they are "eaten" by gauge bosons
 
 We consider one example (the simplest case), a Lagrangian of a complex scalar field $\phi=\frac{1}{\sqrt{2}}(\phi_1+i\phi_2)$, that is a function of 
the coordinate $x$,
 \begin{equation}
 \label{Lagrangian U(1)}
 \mathcal{L}=\partial^{\mu}\phi^*\partial_{\mu}\phi-V(\phi),
 \end{equation}
with a potential of the form
 \begin{equation}
 \label{form of V} 
 V(\phi)= {\mu}^2\phi^*\phi+ \lambda({\phi^*}\phi)^2=\mu^2|\phi|^2+\lambda|\phi|^4.
 \end{equation}
 This Lagrangian has a global $U(1)$ symmetry describing rotation in the complex plane. This means that the Lagragian is invariant 
under global gauge transformations: $\phi\rightarrow\tilde{\phi}=e^{i\alpha}\phi$.
 We know that the vacuum state is the state with minimum potential hence when the parameter $\mu^2>0$, the minimum energy reaches
at $\phi=0$ and the vacuum expectation value for the field $\phi_0=\langle0|\phi|0\rangle=0$. If now $\mu^2<0$, the minimum energy 
no longer corresponds to a unique value of $\phi$. In this case the fields acquire the vacuum expectation value at
\begin{equation}
 |\phi_0|^2=-{\dfrac{\mu^2}{2\lambda}}=\dfrac{v^2}{2}.
\end{equation}      
%With $v^2=-\frac{\mu^2}{\lambda}=\phi^2_{01}+\phi^2_{02}$ is vacuum expectation value. 
We find that the energy is degenerate with 
the minimum at all point on a circle in the $(\phi_1,\phi_2)$ plane with radius $v$. Quantum field theory demands that the vacuum to be unique so that perturbation expansions must be calculated around that point. So, let us choose a minimum on the circle and develop 
the theory around this minimum. This choice leads us to the breaking of the symmetry. A theory where the vacuum has less symmetry than 
the Lagrangian is called a theory with spontaneous symmetry breaking. For simplicity, we can take the real scalar field with non-zero 
expectation value ${\langle}0|\phi_1|0\rangle=v$ and the imaginary part, ${\langle}0|\phi_2|0\rangle=0$. Then, we parametrize $\phi$ 
with two real fields $\theta(x)$ and $\eta(x)$ as
 \begin{equation}
\label{phi}
 \phi(x)=\dfrac{1}{\sqrt{2}}(v+\theta(x)+i\eta(x)).
 \end{equation}
 Inserting \eqref{phi} into \eqref{Lagrangian U(1)} we have:
 \begin{equation}
 \mathcal{L}=\frac{1}{2}(\partial_{\mu}\eta)^2+\frac{1}{2}(\partial_{\mu}\theta)^2+\mu^2\eta^2+ \text{const} + \text{cubic and quartic 
terms in}~ \sigma, \eta .
 \end{equation}
 The third term has the form of a mass term ($-\frac{1}{2}m^2_{\eta}{\eta}^2$) for the $\eta$-field. Thus, this Lagrangian describes 
one real scalar particle $\eta$ with mass $m_{\eta}= \sqrt{-2{\mu}^2} $ and one real massless scalar particle, called Goldstone 
boson $\theta$. Here, the number of Goldstone boson $=1 =$ number of generators which breaks the invariance of the vacuum. 
This reflects the content of Goldstone theorem in a simplest case.

To continue, we research the case which gauge group is $SU(2)$. In this case, the field is a doublet, $\phi=\begin{pmatrix}
\phi_1\\
\phi_2
\end{pmatrix}$,
where, $\phi_1=(\varphi_1+i\varphi_2)/{\sqrt{2}}$, and $\phi_2=(\varphi_3+i\varphi_4)/{\sqrt{2}}$. Then, the global $SU(2)$ invariant Lagrangian as
\begin{equation}
\label{Lagrangian SU(2)}
\begin{split}
\mathcal{L}& =\partial_{\mu}\phi^{+}\partial^{\mu}\phi-V(\phi^{+}\phi),\\
V&= -\mu^2\phi^{+}\phi+\lambda(\phi^{+}\phi)^2,
\end{split}
\end{equation}
with $\mu^2>0$. Now we parametrize as
\begin{equation}
\label{parametrize scalar field}
\phi=\dfrac{1}{\sqrt{2}}e^{i\tau^{i}\xi^i(x)/2\nu}\begin{pmatrix}
0\\
\nu+H(x)
\end{pmatrix},
\end{equation}
where the H(x) and $\xi^i(x)$ are the saclar fields, the VEV $\nu$ is defined as $\varphi^2_{01}+\varphi^2_{02}+\varphi^2_{03}+\varphi^2_{04}\equiv\nu^2=\dfrac{\mu^2}{\lambda}$ or $(\phi^{+}\phi)_0\equiv|\phi^2_0|=\dfrac{\nu^2}{2}=\dfrac{\mu^2}{2\lambda}$. Then, we have
\begin{equation}
\partial_{\mu}\phi=\dfrac{1}{\sqrt{2}}e^{i\tau^i\xi^i/2{\nu}}\lbrace\begin{pmatrix}
0\\
\partial_{\mu}H
\end{pmatrix}
+\dfrac{i}{\nu}\dfrac{\tau^i}{2}\partial_{\mu}\xi^i\begin{pmatrix}
0\\
\nu+H
\end{pmatrix}
{\rbrace},
\end{equation}
we get:
\begin{equation}
\begin{split}
\mathcal{L}& =\dfrac{1}{2}[\begin{pmatrix}
0& \partial_{\mu}H
\end{pmatrix}
-\dfrac{i}{\nu}\partial_{\mu}\xi^i\begin{pmatrix}
0& \nu+H
\end{pmatrix}
\dfrac{\tau^i}{2}]\\
&\times\begin{pmatrix}
0\\
\partial^{\mu}H
\end{pmatrix}
+\dfrac{i}{\nu}\dfrac{\tau^j}{2}\partial^{\mu}\xi^j\begin{pmatrix}
0\\
\nu+H
\end{pmatrix}]
-V((\nu+H)^2),
\end{split}
\end{equation}
using the relation $\tau^i\tau^j=\delta^{ij}+i\varepsilon_{ijk}\tau^k$, we have
\begin{equation}
\mathcal{L}=\dfrac{1}{2}\partial_{\mu}H\partial^{\mu}H+\dfrac{1}{8\nu^2}\partial_{\mu}\xi^i\partial^{\mu}\xi^i(\nu+H)^2-V((\nu+H)^2).
\end{equation} 
From this Lagrangian, we obtain that the 3 fields $\xi^i$ $(i=1,2,3)$ have no mass terms. They are 3 massless Goldstone bosons, while H is massive with mass $m_H=\sqrt{2\mu^2}$. In this example, number of Goldstone is $3$ $\xi^i$, $i=(1,2,3)$ corresponding $3$ generators of $SU(2)$ group  
 
 \subsection{Higgs mechanism}
 
 In this section, the first, we consider the simplest case: $U(1)$ gauge symmetry. First we must make \eqref{Lagrangian U(1)} invariant under 
a $ U(1)$ local gauge transformation,
 \begin{equation}
    \phi \rightarrow e^{i{\alpha}(x)}\phi.
    \end{equation}   
Because of the gauge invariance, we replace the normal derivative $\partial_{\mu}$ by the covariant derivative  
 \begin{equation}
  D_{\mu}=\partial_{\mu}-ieA_{\mu},
  \end{equation}
where the gauge field $A_{\mu}$ transforms as
  \begin{equation}
\label{gf}
   A_{\mu} \rightarrow A_{\mu}+\dfrac{1}{e}\partial\alpha(x).
   \end{equation} 
The gauge invariant Lagrangian is thus
\begin{equation}
\label{Lagrangian scalar QED}
\mathcal{L}=(\partial^{\mu}+ieA^{\mu})\phi^*(\partial_{\mu}-ieA_{\mu}){\phi}-\mu^2\phi^*{\phi} -\lambda(\phi^*\phi)^2 -
\dfrac{1}{4}F_{\mu\nu}F^{\mu\nu}. 
\end{equation}
If $\mu^2>0$ we have the QED Lagrangian of the charged scalar particle of mass $\mu$. But now taking $\mu^2<0$ we have 
generated masses by spontaneous symmetry breaking. 
%The minimum potential V as $|\phi_0|^2=\frac{v^2}{2}=-\frac{\mu^2}{2\lambda}$. 
If we parametrize the field $\phi(x)$ as 
\begin{equation}
\phi(x)=\dfrac{1}{\sqrt{2}}(v+\eta(x))e^{-i\theta(x)/{v}},
\end{equation}
where $\eta(x)$ and $\theta(x)$ are real fields. We define the "unitary gauge" $\alpha(x)=\theta(x)/{v}$, 
\begin{equation}
\label{transformation with unitary gauge}
\begin{split}
\phi(x) &\rightarrow \phi'(x) = e^{i\theta(x)/{v}}\phi(x)=\dfrac{1}{\sqrt{2}}(v+\eta(x))\\
A_{\mu}(x) &\rightarrow B_{\mu}(x) =A_{\mu}(x)-\dfrac{1}{ev}\partial_{\mu}\theta(x).
\end{split}
\end{equation}
Use this unitary gauge transformation, we have
\begin{equation}
D_{\mu}\phi(x) \rightarrow D'_{\mu}\phi'(x)=(\partial_{\mu}-ieB_{\mu})\dfrac{1}{\sqrt{2}}({v}+{\eta}(x)),
\end{equation}
and
\begin{equation}
\label{strength tensor}	
F_{\mu\nu}(A)=\partial_{\mu}A_\nu-\partial_\nu A_{\mu} \rightarrow F_{\mu\nu}(B)=\partial_{\mu}B_\nu-\partial_\nu B_{\mu}.
\end{equation} 
Here we can easily show $F_{\mu\nu}(B)=F_{\mu\nu}(A)$ (gauge invariant). Substituting \eqref{transformation with unitary 
gauge} $\backsim$ \eqref{strength tensor} to \eqref{Lagrangian scalar QED} we have
\begin{equation}
\begin{split}
\mathcal{L}& = \dfrac{1}{2}|\partial_{\mu}\eta-ieB_{\mu}(v+\eta)|^2-\dfrac{\mu^2}{2}(v+\eta)^2\\
&-\dfrac{\lambda}{4}(v+\eta)^4-\dfrac{1}{4}F_{\mu\nu}(B)F^{\mu\nu}(B)\\
& = \dfrac{1}{2}\partial_{\mu}\eta\partial^{\mu}\eta-\mu^2\eta^2-\dfrac{1}{4}F_{\mu\nu}(B)F^{\mu\nu}(B) 
+\dfrac{1}{2}(ev)^2B_{\mu}B^{\mu}\\
& +\dfrac{1}{2}e^2B_{\mu}B^{\mu}\eta(\eta+2v)-{\lambda}v\eta^3-\dfrac{\lambda}{4}\eta^4 .
\end{split}
\end{equation} 
We can see that this Lagrangian describes a massive vector boson $B$ with mass $m_B=ev$ and a massive scalar $\eta$ with 
mass $m_{\eta}=\sqrt{2\mu^2}$. Here the boson Goldstone $\theta(x)$ were eaten by the gauge boson $B$ and became the 
longitudinal component of it. Note that in the Higgs mechanism the degree of freedom is conserved: starting from $2$ real 
scalar fields ($\eta$, $\theta$) and the massless vector fields $A_{\mu}$ we finally got one real massive scalar field 
$\eta$ and one massive vector boson $B_{\mu}$ with $3$ degree of freedom. \cite{Martin84}

To continue, we consider the $SU(2)$ model. The gauge invariant Lagrangian with $SU(2)$ symmetry is given by
\begin{equation}
\mathcal{L}=(D_{\mu}\phi)^+(D^{\mu}\phi)-\dfrac{1}{4}F^i_{\mu\nu}F^{i\mu\nu}-V(\phi^+\phi),
\end{equation}
with
\begin{equation}
\begin{split}
D_{\mu}\phi&= (\partial_{\mu}-ig\dfrac{\tau^i}{2}A^i_{\mu})\phi, (i=1,2,3)\\
F^i_{\mu\nu}&= \partial_{\mu}A^i_{\nu}-\partial_{\nu}A^i_{\mu}+g\varepsilon_{ijk}A^i_{\mu}A^j_{\nu},\\
V(\phi^{+}\phi)&= -\mu^2\phi^+\phi+\lambda(\phi{^+}\phi)^2. (\mu^2>0)
\end{split}
\end{equation}
Now , we parametrize the field $\phi(x)$ as
\begin{equation}
\phi(x)=\dfrac{1}{\sqrt{2}}e^{i\tau^i\xi^i(x)/2\nu}\begin{pmatrix}
0\\
\nu+H(x)
\end{pmatrix},
\end{equation}
with H(x) and $\xi^i(x)$ is the real fields $(i=1,2,3)$, then we use the unitary gauge, define the new fields as
\begin{equation}
\begin{split}
\phi(x)& \rightarrow \phi'(x)=U(x)\phi(x)=\dfrac{1}{\sqrt{2}}\begin{pmatrix}
0\\
\nu+H(x)
\end{pmatrix},\\
A^i_{\mu}& \rightarrow B^i_{\mu}=U(x)A^i_{\mu}U^{-1}- \dfrac{i}{g}(\partial_{\mu}U)U^{-1},
\end{split}
\end{equation}
with
\begin{equation}
U(x)=e^{-i\tau^i\xi^i(x)/2\nu}.
\end{equation} 
This transformation leads to 
\begin{equation}
\begin{split}
D_{\mu}\phi& \rightarrow (D_{\mu}\phi)'=(\partial_{\mu}-ig\dfrac{\tau^i}{2}B^i_{\mu})\dfrac{1}{\sqrt[2]}\begin{pmatrix}
0\\
\nu+H(x)
\end{pmatrix},\\
F^i_{\mu\nu}(A)F^{i\mu\nu}(A)& \rightarrow F^i_{\mu\nu}(B)F^{i\mu\nu}(B)=F^i_{\mu\nu}(A)F^{i\mu\nu}(A),
\end{split}
\end{equation}
with
$F^i_{\mu\nu}(B)=\partial_{\mu}B^i_{\nu}-\partial_{\nu}B^i_{\mu}+g\varepsilon_{ijk}B^j_{\mu}B^k{\nu}$.

Then the Lagrangian becomes
\begin{equation}
\mathcal{L}= (D_{\mu}\phi)'^+(D^{\mu}\phi)'-\dfrac{1}{4}F^i_{\mu\nu}(B)F^{i\mu\nu}(B)+\mu^2(\phi'^+\phi')-\lambda(\phi'+\phi')^2.
\end{equation}


\section{GWS model}
\subsection{Lagrangian in GWS model}

Glashow-Weinberg-Salam (GWS) model is the electroweak part of the standard model (SM). Another part of the SM is the QCD. 
Here, between this thesis, we only consider the GWS model. In the GWS model, the gauge group is $SU(2)_L \times U(1)_Y$, 
where $Y$ is weak-hypercharge and $Q=T^3+\dfrac{Y}{2}$, $T^3$ is the third generations of $SU(2)_L$. Why $SU(2)_L$? Because 
the weak interactions are parity violating, only left-handed fermions, not right-handed ones, participate in the interactions. 
Why $U(1)_Y$ not $U(1)_{em}$? To answer this question, we must start with the charges and the currents. In quantum field theory, 
the relation of the charge $Q^i$ and the current $J^i_{\mu}$ is given as 
\begin{equation}
\label{relation charge-current}
Q^i={\int}J^i_0(x)d^3x.
\end{equation}
With group $U(1)_{em}$ the charge is electric charge, and the electromagnetic currents for a fermion with charge $Q$ is
\begin{equation}
J^{em}_{\mu}(x)=\overline{\psi}\gamma_{\mu}Q\psi,
\end{equation}
where $Q$ is the charge operator of fermion $\psi$. For an electron, the eigenvalue of $Q$ is $Q = -1$.
For $SU(2)_L$ group the weak currents are
\begin{equation}
J^i_{\mu}(x)=\overline{L}\gamma_{\mu}T^iL=\overline{L}\gamma\dfrac{\tau^i}{2}L
\end{equation}
with $\tau^i$ being Pauli matrices, and $L$ is the left-handed part of one fermion-doublet, for example, with the first 
generation of lepton we have
\begin{equation}
   L \equiv \begin{pmatrix}
   \nu_{eL}\\
   e_{L}
   \end{pmatrix}
\text{and we also consider }  R=e_R.
\end{equation}   
The charge (generator of $U(1)_{em}$) for an electron (for example) is $Q$:
\begin{equation}
Q=\int J^{em}_0(x)d^3x=-\int\overline{e}\gamma_{0}ed^3x=-\int(e^+_Le_L+e^+_Le_R)d^3x,
\end{equation}
and the weak charge is $T^i$:
\begin{equation}
T^i=\int J^i_0(x)d^3x.
\end{equation}
We can proof that, the generator $Q$ do not commute with $T^i$, hence $U(1)_{em}$ and $SU(2)_L$ cannot be 
simultaneous symmetries of the model.
This is the answer of the question why we $U(1)_Y$ is used. 

An $SU(2)_L \times U(1)_Y$ - gauge invariant Lagrangian have the form
\begin{equation}
\label{Lagragian GWS}
\mathcal{L}_{GWS}= \mathcal{L}_F + \mathcal{L}_G +  \mathcal{L}_s +  \mathcal{L}_Y.
\end{equation}
We will explain each term of this Lagragian. The first term 
$$\mathcal{L}_F=\overline{L}i\gamma^{\mu}(\partial_{\mu}-ig\dfrac{\overrightarrow{\tau}}{2}\overrightarrow{A}_{\mu}
+\dfrac{i}{2}g'B_{\mu})L+\overline{R}i{\gamma}^{\mu}(\partial_{\mu}
+ig'B_{\mu})R$$ 
is the fermion Lagragian, where $A^i_{\mu}$ ($i=1,2,3$) and $B_{\mu}$ are gauge boson fields associated with 
$SU(2)_L$ and $U(1)_Y$, respectively, while $g$ and $g'$ are gauge 
coupling contants corresponding to $SU(2)_L$ and $U(1)_Y$, respectively. Here the explicit forms of the covariant 
derivates for $L$ and $R$ come out from the general form
\begin{equation}
D_{\mu} = \partial_{\mu}-ig\dfrac{\overrightarrow{\tau}}{2}\overrightarrow{A}_{\mu}-ig'\dfrac{Y}{2}B_{\mu}
 \end{equation} 
 by taking account of $Y = -1$ for $L$ and $Y = -2$ for $R$. Here the fermion mass term do not appear 
 because it violates gauge invariant. Therefore, all fermions are massless at the theory before spontaneous 
breaking.
 
 The second term, 
 \begin{equation}
 \mathcal{L}_G=-\dfrac{1}{4}F^i_{\mu\nu}F^{i\mu\nu}-\dfrac{1}{4}B_{\mu\nu}B^{\mu\nu}, 
 \end{equation}
  with
 \begin{equation}
 \begin{split}
 F^i_{\mu\nu}& = \partial_{\mu}A^i_\nu-\partial_\nu A^i_{\mu}+g\epsilon_{ijk}A^i_{\mu}A^k_\nu \\
 B_{\mu\nu}& = \partial_{\mu}B_\nu -\partial_\nu B_{\mu},
 \end{split}
 \end{equation}
where $F^i_{\mu\nu}$ ($i=1,2,3$) and $B_{\mu\nu}$ are field strength tensors of the gauge fields corresponding to 
$SU(2)_L$ and $U(1)_Y$ is the kinetic term of the gauge fields. Here, the gauge fields are also massless. However, 
in reality, all particles (except photon) are massive. To make fermions and bosons massive, we need 
the spontaneous gauge symmetry breaking and the Higgs mechanism. Since we are living in a $U(1)_{em}$ 
symmetric world corresponding to massless photon, the symmetry breaking could be  
 \begin{equation}
  SU(2)_L \times U(1)_Y \rightarrow U(1)_{em}.
  \end{equation}
  
To realize symmetry breaking and Higgs mechanism, we introduce scalar fields, called Higgs bosons. We start with 4 
gauge bosons (3 associated with $SU(2)_L$ and 1 with $U(1)_Y$), and we have in final one massless 
photon associated with $U(1)_{em}$, we need at least 4 degrees of freedom. The simplest example of such scalars 
is an $SU(2)$ doublet of 2 complex scalar fields which have $Y_{\phi}=+1$,
\begin{equation}
\label{complex scalar fields}
\phi= \begin{pmatrix}
\varphi^{+}\\
\varphi^0
\end{pmatrix},
\end{equation}
where $\varphi^{+}$ and $\varphi^0$ are positively charged and neutral complex scalar fields. The third terms 
of \eqref{Lagragian GWS} is given by 
\begin{equation}
\mathcal{L}_s=(D_{\mu}\phi)^{+}(D^{\mu}\phi)-V(\phi^{+}\phi),
\end{equation}
with 
\begin{equation}
D_{\mu}=(\partial_{\mu}-ig\dfrac{\overrightarrow{\tau}}{2}\overrightarrow{A}_{\mu}-\dfrac{i}{2}g'B_{\mu})\phi .
    \end{equation}
The potential $V(\phi^{+}\phi)$ is given by
\begin{equation}
\label{potentiel scalaire}
    V(\phi^{+}\phi)=\mu^2{\phi}^{+}\phi+\lambda(\phi^{+}\phi)^2,
    \end{equation} 
where $\mu^2$ and $\lambda$ are real constant parameters, and $\lambda >0$ to ensure stability of the vacuum.

The final term of \eqref{Lagragian GWS} is added "by hand". It describe interactions between scalars and fermions, 
called Yukawa interaction term, which is $SU(2)_l \times U(1)_Y$ gauge invariant and is to provide the fermions 
mass after spontaneous symmetry breaking,
\begin{equation}
\mathcal{L}_Y= -G_e(\overline{L}{\phi}R+\overline{R}\phi^{+}L)+h.c,
\end{equation}    
where $G_e$ is an Yukawa coupling constant which cannot be determined within the GWS model itself.
 
\subsection{Spontaneous breaking in GWS model}

From the previous section, we know that the potential $V(\phi^{+}\phi)$ of \eqref{potentiel scalaire} with positive 
$\lambda$ and negative $\mu^2$ has a minimum at 
\begin{equation}
\phi^{+}\phi=|\phi|^2=\dfrac{v^2}{2} \text{with} v=\sqrt{\dfrac{-\mu^2}{\lambda}}.
\end{equation}
The vacuum expectation value of the scalar doublet $\phi$ is 
\begin{equation}
\phi_0=\langle{0}|\phi|0\rangle=\begin{pmatrix}
0\\
v/{\sqrt{2}}
\end{pmatrix}.
\end{equation}
Now, we parametrize the scalar doublet in terms of the fields denoting the shifts from the vacuum state $\phi_0$
\begin{equation}
\phi=\begin{pmatrix}
\varphi^{+}\\
\varphi^0
\end{pmatrix}
=e^{i\overrightarrow{\tau}.\overrightarrow{\xi}/2v}\begin{pmatrix}
0\\
(v+H)/{\sqrt{2}}
\end{pmatrix}
\end{equation}
Here the original $2$ complex scalar fields in \eqref{complex scalar fields} are replaced by $4$ real fields $\xi_i$ $(i=1,2,3)$ 
and $H$, where $\xi_i$ are the Goldstone bosons being absorbed into the longtudinal components of $W^{\pm}$ and $Z^0$ 
bosons and $H$ is a Higgs boson. Each of these fields has a zero vacuum expectation value. We choose the "unitary gauge", use the 
unitary $SU(2)$ transformation $ U(\xi)=e^{-i\overrightarrow{\tau}.\overrightarrow{\xi}/2v} $ and rewrite Lagrangian scalar 
(after the unitary $SU(2)$ transformation $ \phi \rightarrow \phi'= U(\xi)\phi$, $ L \rightarrow L'=U(\xi)L $, 
$ \overrightarrow{A_{\mu}} \rightarrow \overrightarrow{A'_{\mu}}=U(\xi)\overrightarrow{A_{\mu}}U(\xi)^{-1}-
\dfrac{i}{g}(\partial_{\mu}U(\xi))U^{+}(\xi) $, $R'=R$, $ B'_{\mu}=B_{\mu} $) we have
\begin{equation}
\mathcal{L}_s = (D_{\mu}\phi)'(D^{\mu}\phi)'-V(\phi'^{+}\phi'),
\end{equation}
with
\begin{equation}
\label{derive covariant apres unitary gauge}
\begin{split}
 (D_{\mu}\phi)'& = (\partial_{\mu}-ig\dfrac{\overrightarrow{\tau}}{2}\overrightarrow{A'}_{\mu}\dfrac{i}{2}g'B'_{\mu})\phi'\\
 &=(\partial_{\mu}-ig\dfrac{\overrightarrow{\tau}}{2}\overrightarrow{A'}_{\mu}\dfrac{i}{2}g'B'_{\mu})\dfrac{1}{\sqrt{2}}(v+H)\chi,
 \end{split}
\end{equation}
where
$ \chi=\begin{pmatrix}
0\\
1
\end{pmatrix}. $
The first term of \eqref{derive covariant apres unitary gauge} contain the mass-squared term for weak gauge bosons. Developing it 
we have 
\begin{equation}
\label{lagrangien de la masse}
\begin{split}
\mathcal{L}_{mass}&= \dfrac{v^2}{2}\chi^{+}(g\dfrac{\overrightarrow{\tau}}{2}.\overrightarrow{A}'_{\mu}+\dfrac{g'}{2}B'_{\mu})
(g\dfrac{\overrightarrow{\tau}}{2}.\overrightarrow{A}'^{\mu}+\dfrac{g'}{2}B'^{\mu})\chi\\
&=\dfrac{v^2}{8}(g^2\overrightarrow{A}'_{\mu}.\overrightarrow{A}'^{\mu}+g'^2B'_{\mu}B'^{\mu}-2gg'B'_{\mu}A'^{3\mu})\\
&=\dfrac{v^2}{8}(g^2A'^1_{\mu}A'^{1\mu}+g^2A'^2_{\mu}A'^{2\mu}+(gA'^3_{\mu}-g'B'_{\mu})^2).
\end{split}
\end{equation}
To calculate from the $ 1^{st} $ line to the $ 2^{nd} $ line , we used the formula $ \tau^i{\tau^j}= \delta^{ij}+i\epsilon^{ijk}\tau^k $. 
Defining
\begin{equation}
\label{definition W}
 W^{\pm}_{\mu}=\dfrac{A'^1_{\mu}\mp iA'^2_{\mu}}{\sqrt{2}},
 \end{equation} 
we can rewrite the sum of the $1^{st}$ and $2^{nd}$ terms of \eqref{lagrangien de la masse} as $\dfrac{1}{4}g^2{v}^2W^{+}_{\mu}W^{-\mu}$. 
It means the bosons $W^{\pm}$ have the mass 
\begin{equation}
\label{masse de W}
M_W=\dfrac{1}{2}g{v}.
\end{equation}

The third term can be written as
\begin{equation}
\dfrac{v^2}{8}\begin{pmatrix}
A'^3_{\mu}& B'_{\mu}
\end{pmatrix}
\begin{pmatrix}
g^2& -gg'\\
-gg'& g'^2
\end{pmatrix}
\begin{pmatrix}
A'^{3\mu}\\
B'^{\mu}
\end{pmatrix}.
\end{equation}
Diagonalizing it we obtain 
\begin{equation}
\begin{pmatrix}
A'^3_{\mu}& B'_{\mu}
\end{pmatrix}
\begin{pmatrix}
g^2& -gg'\\
-gg'& g'^2
\end{pmatrix}
\begin{pmatrix}
A'^{3\mu}\\
B'^{\mu}
\end{pmatrix}
=\begin{pmatrix}
Z_{\mu}& A_{\mu}
\end{pmatrix}
\begin{pmatrix}
g^2+g'^2& 0\\
0&        0
\end{pmatrix}
\begin{pmatrix}
Z^{\mu}\\
A^{\mu}
\end{pmatrix},
 \end{equation} 
with 
\begin{equation}
\label{diagonal}
\begin{pmatrix}
Z_{\mu}\\
A_{\mu}
\end{pmatrix}
=\begin{pmatrix}
cos{\theta_{W}}& -sin{\theta_{W}}\\
sin{\theta_{W}}& cos{\theta_W}
\end{pmatrix}
\begin{pmatrix}
A'^3_{\mu}\\
B'_{\mu}
\end{pmatrix},
\end{equation}
where $\theta_W$ is called the weak mixing angle, or Weinberg angle, and defined by $tg{\theta}=\dfrac{g'}{g}$. From \eqref{diagonal} we have
\begin{equation}
\label{masse de Z}
M_Z=\dfrac{1}{2}v\sqrt{g^2+g'^2}=\dfrac{M_W}{cos{\theta}_W}.
 \end{equation} 
Note that the quantity $\rho=\dfrac{M_W^2}{M_Z^2cos^2{\theta}_W}$ takes the value $\rho=1$ in the SM.

\chapter{Vacuum Polarization}
\section{Introduction to the vacuum polarization}
In quantum fields theory (QFT), vacuum is treated as a sea consisting of many virtual particle-antiparticle pairs appear and disappear 
immediately. This process in quantum electrodynamics (QED) is described by one-loop diagrams, diagrams of self-energies of photon. In general, 
with gauge fields it is describes by diagrams of self energies of gauge fields. Because of the time is very short, the Heisenberg uncertainty 
principle allows violation of conservation during that time. As a consequence, the invariant momentum of the fermions in the loop is allowed 
to be higher than the invariant momentum of the propagator. The loop is called to be off mass shell. For this case, $q^2<4m^2$, with $m$ the mass 
of the fermions inside the loop and $q^2$ the invariant momentum of the photon propagator. When the loop is on mass shell, $q^2>4m^2$.

A vacuum polarization changes a gauge propagator (the first order contribution in perturbation theory to the photon propagator in the case of QED). 
For example, in QED, the photon propagator when we calculate contribution of the vacuum polarization (from the Feynman rule) changes as
\begin{equation}
\dfrac{-ig_{\mu\nu}}{q^2+i\epsilon}\Rightarrow\dfrac{-ig_{\mu\nu}}{q^2+i\epsilon}+\dfrac{-ig_{\mu\rho}}{q^2+i\epsilon}\Pi^{\rho\sigma}(q)
\dfrac{-ig_{\sigma\nu}}{q^2+i\epsilon}, 
\end{equation}
with $\Pi^{\mu\nu}(q)$ the contribution of the loop. To get the first order contribution in perturbation theory, we need to calculate this 
contribution.

Because the uncertainty principle when we calculate the loop corrections, we must integrate over the internal loop momentum, extending from 
$-\infty$ to $+\infty$. But when  the Feynman rules are applied onto this diagram, there will be two factors of $q^{-2}$ in the amplitude 
caused by the two fermions inside the loop. A four dimension integral like $\int\frac{d^4q}{q^4}$ will be divergent. Contribution of loop 
is infinite contraction, hence to calculate loop corrections we must use special techniques: regularization and renormalization to remove 
divergent part of loop. There are 3 methods of regularization: dimensional regularization and Pauli-Villars regularization. 
Between this master thesis (hence the limits of number of pages), in this chapter, we only calculate loop corrections to vacuum polarization 
on QED and only method of dimension regularization. Result of calculation by Pauli-Villars regularization is written in result of loop 
contribution because they are useful to calculations in chapter 3.
 
\section{Calculation of the vacuum polarization by the method of dimension regularization }
Here, we work with a space of dimension $d=4-\epsilon$ to separate the divergent part of an amplitude. Divergent parts in this method 
go to infinity as $\frac{1}{\epsilon}$ when $\epsilon\rightarrow0$. In a $d$-dimension space the algebra of Dirac matrices is the following
\begin{equation}
\begin{split}
\lbrace\gamma^{\mu},\gamma^{\nu}\rbrace& =g^{\mu\nu}=diag(1,-1,-1,...,-1),\\
\gamma^{\mu}\gamma_{\mu}& =d, ~ \gamma_{\mu}\gamma_{\nu}\gamma^{\nu}=(2-d)\gamma_{\nu},\\
Tr(\gamma)&=0, ~ TrI=d, ~ Tr{\gamma_{\mu}\gamma_{\nu}}=dg_{\mu\nu},\\
Tr{\gamma_{\alpha}\gamma_{\beta}\gamma_{\mu}\gamma_{\nu}}&=d(g_{\alpha\beta}g_{\mu\nu}-g_{\alpha\mu}g_{\beta\nu}+g_{\alpha\nu}g_{\beta\mu}).
\end{split}
\end{equation}
From the Feynman rules (in $d$-dimension, the vertex of electromagnetic interaction $ -ie\gamma^{\mu} $ is replaced by \cite{Ryder96}, \cite{Derendinger2001}, 
$ -ie\mu^{2-\frac{d}{2}}\gamma^{\mu} $, to regularize the dimension in Lagrangian, $ \mu $ disappear when $ \epsilon \rightarrow 0 $) we have 
\begin{equation}
\begin{split}
i\Pi^{\mu\nu}(q)& =-\mu^{4-d}\int\dfrac{d^dp}{(2\pi)^d}tr[(-ie\gamma^\mu)\dfrac{i(\slashed k+m)}{k^2-m^2}(-ie\gamma^\nu)
\dfrac{i(\slashed k-\slashed q+m)}{(k-q)^2-m^2}]\\
&=-e^2\mu^{4-d}\int\dfrac{d^dk}{(2\pi)^d}\dfrac{N^{\mu\nu}(k,q)}{(k^2-m^2)((k-q)^2-m^2)}.
\end{split} 
\end{equation}
Calculating traces of Dirac matrices in $d$-dimension we have
\begin{equation}
\Pi^{\mu\nu}(q)=ie^2d{\mu}^{4-d}\int\dfrac{d^dk}{(2\pi)^d}\dfrac{2k^{\mu}k^{\nu}-(k^{\mu}q^{\nu}+k^{\nu}q^{\mu})
-g^{\mu\nu}(k^2-kq+q^2)}{(k^2-m^2)((k-q)^2-m^2)}.
\end{equation}
Using the Feynman parametrization \cite{Vlijm2010}
\begin{equation}
\dfrac{1}{AB}=\int^1_0\dfrac{dx}{[xA+(1-x)B]^2},
\end{equation}
where 
\begin{equation}
\begin{split}
A &=k^2-m^2,\\
B &=(k-q)^2-m^2,
\end{split}
\end{equation}
we obtain
\begin{equation}
\label{l'expression de la polarization du vide}
\begin{split}
\Pi^{\mu\nu}(q)&=ie^2\mu^{4-d}d\int^1_0dx{\int}\dfrac{d^dk}{(2\pi)^d}\lbrace\dfrac{2k^{\mu}k^{\nu}}{[k^2-m^2+q^2x(1-x)]^2}\\
&-\dfrac{2x(1-x)[q^{\mu}q^{\nu}-g^{\mu\nu}q^2]}{[k^2-m^2+{q^2}x(1-x)]^2}-\dfrac{g^{\mu\nu}}{k^2-m^2+q^2x(1-x)}\rbrace .
\end{split}
\end{equation}
When we calculate integrals in a $d$-dimensional space like \eqref{l'expression de la polarization du vide}, we use the following formulas
\begin{equation}
\label{l'intergrale 1}
\int{d^dk}\dfrac{1}{(k^2+2kq-m^2)^n}=\dfrac{i(-1)^n(\pi)^{d/2}}{\Gamma(n)(m^2+q^2)^{n-d/2}}\Gamma(n-\dfrac{d}{2}),
\end{equation}
\begin{equation}
\label{l'intergrale 2}
\int{d}^dk\dfrac{k^{\mu}}{(k^2+2kq-m^2)^n}=\dfrac{-i(-1)^n\pi^{d/2}}{\Gamma(n)(m^2+q^2)^{n-d/2}}q^{\mu}\Gamma(n-\dfrac{d}{2})
\end{equation}
and
\begin{equation}
\label{l'intergrale 3}
\begin{split}
\int{d}^dk\dfrac{k^{\mu}k^{\nu}}{(k^2+2kq-m^2)^n}&=\dfrac{i(-1)^n\pi^{d/2}}{\Gamma(n)(m^2+q^2)^{n-d/2}}[q^{\mu}q^{\nu}\Gamma(n-\dfrac{d}{2})\\
&-\dfrac{1}{2}g^{\mu\nu}(q^2+m^2)\Gamma(n-d/2-1)],
\end{split}
\end{equation}
where the gamma function is given as follows
\begin{equation}
\label{fonction gamma}
\begin{split}
\Gamma(-n+\epsilon)&=\dfrac{(-1)^n}{n!}[\dfrac{1}{\epsilon}+\psi_1(n+1)+0(\epsilon)]\\
\psi_1(n+1)&=11+\dfrac{1}{2}+\dfrac{1}{3}+...+\dfrac{1}{n}-\gamma
\end{split}
\end{equation}
with $ \gamma=0,577 $ being Euler constant. The following approximation is also used \cite{HoKimQuang98} 
\begin{equation}
\label{formule d'approximation}
a^{\epsilon}\approx 1+\epsilon lna .
\end{equation}
Since the first and the third term in \eqref{l'expression de la polarization du vide} are canceled, 
we have \cite{Zee2003}
\begin{equation}
\begin{split}
\Pi^{\mu\nu}(q)&=-2ie^2d\dfrac{\mu^{4-d}}{(2\pi)^d}\int^1_0x(1-x)(q^{\mu}q^{\nu}-q^2g^{\mu\nu})\times\\
&\times\int{d}^dk\dfrac{1}{[k^2-m^2+x(1-x)q^2]^2}\\
&=(-1)^{d/2}e^2d\dfrac{\mu^{4-d}}{(4\pi)^{d/2}}\int^1_0x(1-x)(q^{\mu}q^{\nu}-q^2g^{\mu\nu})\times\\
&\times\dfrac{\Gamma(2-\frac{d}{2})}{[k^2-m^2+x(1-x)q^2]^{2-d/2}}.\\
\end{split}
\end{equation}
Using \eqref{fonction gamma} and \eqref{formule d'approximation} we get
\begin{equation}
\begin{split}
\Pi^{\mu\nu}(q)&=\dfrac{e^2}{2{\pi}^2}(q^{\mu}q^{\nu}-q^2g^{\mu\nu})(\dfrac{1}{3\epsilon}-\dfrac{\gamma}{6}-ln(4\pi)\\
&-\int^1_0dxx(1-x)ln(\dfrac{m^2-q^2x(1-x)}{{\mu}^2})).
\end{split}
  \end{equation}  
In the Pauli-Villars regularization $\Pi^{\mu\nu}(q)$ has the form 
\begin{equation}
\Pi^{\mu\nu}(q)=(q^{\mu}q^{\nu}-q^2g^{\mu\nu})\dfrac{-q^2}{2\pi^2}\int^1_0dxx(1-x)ln(\dfrac{\Lambda^2}{m^2-x(1-x)q^2}).
\end{equation}
Putting $\Pi^{\mu\nu}(q)=(q^{\mu}q^{\nu}-q^2g^{\mu\nu})\Pi(q^2)$ we have
\begin{equation}
\Pi(q^2)=\dfrac{-q^2}{2\pi^2}\int^1_0dxx(1-x)ln(\dfrac{\Lambda^2}{m^2-x(1-x)q^2}).
\end{equation} 
The Pauli-Villars form will be used in the next chapter. 

\section{The Passarino-Veltman function}

 

\chapter{$S,T,U$ parameters and their interpretation}
\section{A brief introduction of $S,T,U$ parameters}

The Standard Model, especially its electroweak sector, was devised as a renornalisable gauge theory with massive vector 
bosons such as $Z^0$ and $W^\pm$. At tree level, the SM has its characteristic properties and such properties 
have been extensively tested experimentally. Some of the remarkable results are the discovery to predicted neutral current 
processes, the confirmation of the existence of $Z^0$ and $W^\pm$ with their predicted masses. If the scalar boson discovered 
recently by the LHC is the SM Higgs boson, the SM is almost completely tested and confirmed. Let us focus on the gauge sector, 
namely the part of the Lagrangian, which contains gauge bosons. Then we have 3 bare parameters: 2 gauge couplings $g$, $g'$ of 
$SU(2)_L$ and $U(1)_Y$ 
and vacuum expectation value (VEV) of the neutral Higgs boson $v$. What we should do first is to calculate observables 
which have been excellently well-measured experimentally.
\begin{align}
\label{les base parametre}
M_Z& = 91.150(30) GeV \\
G_F& = 1.16637(2)\times10^{-5} (GeV)^{-2} \\
\alpha& = 137.0359895(61)^{-1}
\end{align}    
In the calculation process of one loop corrections, other parameters were discovered. These parameters are important for searching 
heavy particles in Standard Model and New Physics. Between these parameters, 3 good parameters called $S$-,$T$-,$U$ parameters (or 
Peskin-Takechi parameters) are more sensitive to potential New Physics (NP-physics beyond SM) contributions to electroweak 
radiative corrections. We know that, the heavy particles predicted by NP (the $4^{th}$ generation for example) cannot directly appear 
in the external lines of Feynman diagrams for lower energy processes of light particles because they decay very fast. Hence we 
can only obtain effects of NP indirectly by their contributions to the gauge boson self energy, so called oblique electroweak 
corrections. 

The oblique electroweak corrections appear when we research the processes $\bar{f}f\rightarrow\bar{f\prime}f\prime$ with $f$ and 
$f\prime$ are external fermions. Here, we consider weak-interaction processes involving only light fermions as 
external particles since they are the only processes accessible to present day experiments. This restriction has some important 
consequences which simplify our analysis considerably.

Firstly, we can neglect the term proportional to $q^{\mu}q^{\nu}$ in the W and Z propagators. In fact, let us consider the 
lowest order process of $\nu_e+e\rightarrow\nu_e+e$ in the low energy, which occurs through an exchange of $W^\prime$ boson. 
The corresponding Feynman amplitude is given by
 
\begin{equation}
M=-\dfrac{g^2}{2}J^{+\mu}\dfrac{i(-g_{\mu\nu}+\dfrac{q_\mu{q_\nu}}{M^2_W})}{q^2-M^2_W+i\epsilon}J^{-\nu},
\end{equation}
which reduces to
\begin{equation}
M=-i\dfrac{g^2}{M^2_W}J^{+\mu}J^{-}_{\mu}.
\end{equation}
In the limit of $q^2/M^2_W\rightarrow0$, the $q^{\mu}q^{\nu}$ term of the photon propagator has no effect, due to the Ward identity.\cite{Peskin92}
Secondly, we can assume that radiative corrections due to physics beyond standard model appear dominantly through vacuum polarizations 
(oblique corrections) and that vertex corrections and box diagrams (direct corrections) can be neglected.

We start by introducing some notations: $J^{\mu}_Q$,$J^{\mu}_3$, and $J^{\mu}_{\pm}=J^{\mu}_1{\pm}iJ^{\mu}_2$ denote the electromagnetic 
and weak-isospin currents coupling to the electroweak gauge bosons via
\begin{equation}
\label{eq Lagrangian electroweak1}
\mathcal{L}=\frac{e}{\sqrt{2}s}(W^+_{\mu}J^{\mu}_++W^-_{\mu}J^{\mu}_-)+\frac{e}{sc}Z_{\mu}(J^{\mu}_3-s^2J^{\mu}_Q)+eA_{\mu}J^{\mu}_Q,
\end{equation}
where $s=sin{\theta}_W$ and $c=cos{\theta}_W$. The first term of \eqref{eq Lagrangian electroweak1} is charge current Lagrangian 
$\mathcal{L}_{CC}$, and the next terms is neutral current Lagrangian $\mathcal{L}_{NC}$.  We denote the coefficient of $g^{\mu\nu}$ 
in the photon, $Z$ and $W$ propagator by $G_{\gamma\gamma}$, $G_{ZZ}$, and $G_{WW}$, respectively, and that of the photon and $Z$ 
mixing by $G_{{\gamma}Z}$.     
Vacuum polarizations affect the above interactions by modifying the gauge-boson propagator $G_{\gamma\gamma}$, $G_{{\gamma}Z}$, 
$G_{ZZ}$ and $G_{WW}$. The propagator after vacuum polarizations have the form $\dfrac{1}{q^2-M^2_{0B}-\Pi(q^2)}$ for $B=\gamma,Z,W$ 
and $M^2_{0B}$ are the bare masses. This is the reason why they are called "oblique" corrections as opposed to the "direct" vertex 
and box corrections which modify the form of interactions themselves. We define the vacuum polarization functions of the gauge indices 
$(i,j)$, $\Pi_{ij}(q^2)$, where $(i,j)$ =$(11)$, $(22)$, $(33)$, $(3Q)$ and $(QQ)$ by
\begin{equation}
\Pi_{ij}(q^2)=\langle{J^{\mu}_{i},J^{\mu}_{j}}\rangle.
\end{equation}
Form \eqref{eq Lagrangian electroweak1} we can replace the vacuum polarization functions $\Pi_{\gamma\gamma}, \Pi_{{\gamma}Z}, \Pi_{ZZ}$ 
and $\Pi_{WW}$ by gauge indices $(i,j)$ as follows
\begin{align}
\label{eq:tranformations les jauge indices 1}
 \Pi_{\gamma\gamma}& =e^2\Pi_{QQ}\\
 \label{eq:tranformations les jauge indices 2 }
 \Pi_{{\gamma}Z}& =\dfrac{e^2}{sc}(\Pi_{3Q}-s^2\Pi_{QQ})\\
 \label{eq:tranformations les jauge indices 3}   
 \Pi_{ZZ}& =\dfrac{e^2}{s^2c^2}(\Pi_{33}-2s^2\Pi_{3Q}+s^4\Pi_{QQ})\\
 \label{eq:tranformations les jauge indices 4}
 \Pi_{WW}& =\dfrac{e^2}{s^2}\Pi_{11}.
\end{align}
To demonstrate these equation we rewrite equation \eqref{eq Lagrangian electroweak1} as
\begin{equation}
\mathcal{L}=(W^+_{\mu}J^{\mu}_{W+}+W^-_{\mu}J^{\mu}_{W-})+Z_{\mu}J^{\mu}_Z+A_{\mu}J^{\mu}_{\gamma}.
\end{equation}
Let us, for example, demonstrate equation \eqref{eq:tranformations les jauge indices 1}. From the definition we have
\begin{align*}
\Pi_{\gamma\gamma}&=\langle{J^{\mu}_{\gamma}J^{\mu}_{\gamma}}\rangle\\ 
&=e^2\langle{J^{\mu}_QJ^{\mu}_Q}\rangle\\  
&=e^2\Pi_{QQ},
\end{align*}
and similarly with \eqref{eq:tranformations les jauge indices 2 },  \eqref{eq:tranformations les jauge indices 3} and 
\eqref{eq:tranformations les jauge indices 4}. Note that $\Pi_{11}=\Pi_{22}$ due to unbroken symmetry of electromagnetism group.

When we compute one-loop corrections, the values of physics quantities in one loop are different with those at the tree level. 
We denote physics quantities in one loop by attaching a "star" $(*)$. Thus value of these quantities with ``star'' = their value 
of at the tree level + contributions of oblique correction. For example, $\rho$ parameter defined as
\begin{equation}
\label{definition rho}
\rho \equiv \dfrac{M^2_W}{M^2_Zcos^2{\theta}_W}.
\end{equation}
In Standard Model, $\rho=1$ (because $M_W=\frac{M_Z}{cos{\theta}_W}$) at the tree level but when we compute one loop corrections 
we get $\rho_*=1$ + one-loop contribution. The one-loop contributions also contain contributions of New Physics. Hence we consider 
$\Delta\rho=\rho_*-1$ and computing this quantity we have
\begin{equation}
\Delta\rho=\Pi_{11}-\Pi_{33}.
\end{equation}

Because of the functions of vacuum polarization depend on $q^2$, the ``star'' quantities are also functions of $q^2$ and each of 
them has an infinite number of observables as the functions of heavy particle masses appearing as the coefficients of the Taylor 
expansion in terms of $q^2$. Suppose that, we have a vacuum polarization function $\Pi_{ij}(q^2)$ where the indices $(i,j)$ take 
either $SU(2)$ adjoin indices $1,2,3$ or $Q$ corresponding to unbroken $U(1)_{em}$. At the 1-loop level, vertices of the diagrams 
of vacuum polarization are all proportional to dimensionless gauge couplings (at higher loops levels we may have mass depend  
couplings due to the Higgs exchange). In the Taylor expansion it as the power series in $q^2$ or equivalently in $\dfrac{q^2}{M^2}$ 
(where $M$ is a generic mass of intermediate heavy particles),
\begin{equation}
\label{expansion de Taylor}
\Pi_{ij}(q^2)=\Pi_{ij}(0)+\dfrac{d{\Pi}_{ij}}{dq^2}|_{q^2=0}q^2+...
\end{equation}
We treat only the first two terms of this expansion. Thus, at the first sight there seem to exist 8 (=$2\times 4$) parameters, 
corresponding to the $4$ choices of the gauge indices $(i,j)$: $(1,1)=(2,2)$, $(3,3)$, $(3,Q)$,$(Q,Q)$, or equivalently, the choices 
of gauge bosons in external lines of the vacuum polarization diagrams $(W^+,W^-)$, $(Z,Z)$, $(\gamma,\gamma)$, $(Z,\gamma)$. 
Because $\Pi_{3Q}(0)=\Pi_{QQ}(0)=0$ (Ward identity), we thus have $6$ parameters remaining. From these $6$ parameters, tree parameters 
are used for renormalization, in the process to fix the bare parameters in terms of $\alpha,G_F,M_Z$. In other words, there are inputs 
rather than the predictions of the theory. In this way, the outputs of the theory, the genuine predictions of the theory, are in the 
remaining $3$ parameters, which are nothing but the $S,T,U$-parameters we are interested in.

As is clear from the above argument, such three parameters $S,T,U$ should be automatically finite quantities, as they correspond to 
operators, which do not exist in the original Lagrangian: they do not need to be renormalized. We define the following finite 
combinations of the Taylor-series coefficients as new weak-interactions parameters,
\begin{align}
\label{definition S}
{\alpha}S& = 4e^2[\Pi'_{33}(0)-\Pi'_{3Q}(0),]\\
\label{definition T}
{\alpha}T& = \dfrac{e^2}{c^2s^2M^2_Z}[\Pi_{11}(0)-\Pi_{33}(0)],\\
\label{definition U}
{\alpha}U& = 4e^2[\Pi'_{11}(0)-\Pi'_{33}(0)].
\end{align}
We can also rewrite these parameters by choices of gauge bosons in the external lines of the vacuum polarization diagrams, \cite{TASI97},
\begin{equation}
\label{autre facon de S,T,U}
\begin{split}
 {\alpha}S & = 4s^2c^2[\Pi'_{ZZ}(0)-\dfrac{c^2-s^2}{sc}\Pi'_{Z\gamma}(0)-\Pi'_{\gamma\gamma}(0)],\\
 {\alpha}T & = \dfrac{\Pi_{WW}(0)}{M^2_W}-\dfrac{\Pi_{ZZ}(0)}{M^2_Z},\\
 {\alpha}U & = 4s^2[\Pi'_{WW}(0)-c^2\Pi'_{ZZ}(0)-2sc\Pi'_{Z\gamma}(0)-s^2\Pi'_{\gamma\gamma}(0)].
\end{split}
\end{equation}
From the $S,T,U$ parameters we can rewrite the star quantities as \cite{Peskin92} 

\begin{equation}
\begin{split}
\label{quelques quantites etoile}
\dfrac{m^2_{W*}}{m^2_Z}-c^2_0& =\dfrac{{\alpha}c^2}{c^2-s^2}[-\dfrac{1}{2}S+c^2T+\dfrac{c^2-s^2}{4s^2}U],\\
s^2_*(q^2)-s^2_0& = \dfrac{\alpha}{c^2-s^2}(\frac{1}{4}S-s^2c^2T),\\
\rho_*(0)-1& = {\alpha}T,\\
Z_{Z*}(q^2)-1& = \dfrac{\alpha}{4c^2s^2}S,\\
Z_{W*}(q^2)-1& =\dfrac{\alpha}{4s^2}(S+U).
\end{split}
\end{equation}

\section{Calculation of $S,T,U$ parameters and their interpretation}           
Firstly, we can prove that $U$ is unimportant because parameter $U$ only appear in $Z_{W*}$. This means that all 
neutral-current and low-energy observable depend only on $S$ and $T$. In fact, the only accurately measured 
weak-interaction observable that depends on $U$ is $m_W$. In addition, $U$ is often predicted to be very small. 
We assume that a custodial symmetry constrains the $\Pi_{11}(0)$ and $\Pi_{33}(0)$ to be equal to the level of order 
$1\%$ with the difference generated by radiative corrections. In most models, this approximation equality holds for 
all $q^2$, so that $U$ should differ from zero by only a percent of $T$. Hence, we will often add to the assumption 
that $U=0$ and only calculate $S$ and $T$.

From the definition of $S$ and $T$ parameters, we must calculate the vacuum polarization amplitude $\Pi^{\mu\nu}_{33}$, 
$\Pi^{\mu\nu}_{11}$ and $\Pi^{\mu\nu}_{3Q}$. In this master thesis, we only calculate these parameters in the simplest 
case, the case of oblique electroweak corrections of new heavy fermions. As long as we can ignore small elements of the 
quark or the lepton mixing matrices the contributions of new fermions to electroweak processes appear via the simple 
one-loop diagrams. We consider a fermion doublet, for example quark of $4^{th}$ generation, $(t',b')$ with usual 
left-handed coupling to $SU(2)$, hypercharge $Y$, and masses $m_1$, $m_2$ (masses of new generation must be very heavy 
because we have not seen yet particles of this generation). Because we only consider electroweak interaction, we must 
consider the polarization left-right handed of the fermions. Te vacuum polarization amplitude is written as product of 
current-current. \cite{Peskin90} Hence we must calculate the vacuum polarization with left-handed currents and right-handed currents. 
There are $4$ combinations of left and right-handed currents, thus there are 4 vacuum polarization amplitude correspondences 
($ \Pi^{\mu\nu}_{LL} $, $ \Pi^{\mu\nu}_{LR} $, $ \Pi^{\mu\nu}_{RL} $, $ \Pi^{\mu\nu}_{RR}$). 

From the Feynman rules, we have \cite{Peskin95}
\begin{equation}
\begin{split}
\Pi_{LL}^{\mu\nu}&=(-1)\int\dfrac{d^4k}{(2\pi)^4}tr[(i\gamma^{\mu})(\dfrac{1-\gamma^5}{2})\dfrac{i(\slashed k+m)}{k^2-m_1^2}(i\gamma^{\nu})(\dfrac{1-\gamma^5}{2})(\dfrac{i\slashed k+\slashed q+m_2}{(k+q)^2-m_2^2})]\\
&=-\int\dfrac{d^4k}{(2\pi)^4}tr[\gamma^{\mu}\slashed k\gamma^{\nu}(\slashed k+\slashed q)(\dfrac{1+\gamma^5}{2})]\dfrac{1}{(k^2-m_1^2)((k+q)^2-m_2^2)},
\end{split}
\end{equation} 
and
\begin{equation}
\begin{split}
\Pi_{LR}^{\mu\nu}&=(-1)\int\dfrac{d^4k}{(2\pi)^4}tr[(i\gamma^{\mu})(\dfrac{1-\gamma^5}{2})\dfrac{i(\slashed k+m)}{k^2-m_1^2}(i\gamma^{\nu})(\dfrac{1+\gamma^5}{2})(\dfrac{i\slashed k+\slashed q+m_2}{(k+q)^2-m_2^2})]\\
&=-\int\dfrac{d^4k}{(2\pi)^4}tr[\gamma^{\mu}m_1\gamma^{\nu}m_2(\dfrac{1+\gamma^5}{2})]\dfrac{1}{(k^2-m_1^2)((k+q)^2-m_2^2)}.
\end{split}
\end{equation}

Calculating similarly as in chapter $2$ (very complicated) and transforming to Pauli-Villars regularization form we obtain the vacuum polarization\cite{Peskin95}, \cite{Morii2004} 
function
\begin{equation}
\begin{split}
\Pi_{LL}(m^2_1,m^2_2,q^2)&=\Pi_{RR}(m^2_1,m^2_2,q^2)\\
&=-\dfrac{1}{4\pi^2}\int^1_0dxln(\dfrac{\Lambda^2}{M^2-x(1-x)q^2})[x(1-x)q^2\\
&-\dfrac{1}{2}M^2]\\
\Pi_{LR}(m^2_1,m^2_2,q^2)&=\Pi_{RL}(m^2_1,m^2_2,q^2)\\
&=-\dfrac{1}{4\pi^2}\int^1_0dxln(\dfrac{\Lambda^2}{M^2-x(1-x)q^2})\dfrac{1}{2}m_1m_2,
\end{split}
 \end{equation}
where $ M^2=xm^2_1+(1-xm^2_2) $, and $x$ is the Feynman parameter. Writing the vacuum polarization function induced by vector currents as
 \begin{equation}
\Pi_{VV}=\Pi_{L+R,L+R}=\Pi_{LL}+\Pi_{LR}+\Pi_{RL}+\Pi_{RR}=2(\Pi_{LL}+\Pi_{LR})
  \end{equation} 

we get the vacuum polarization functions $ \Pi_{11} $, $ \Pi_{33} $, $ \Pi_{3Q} $ and $ \Pi_{QQ} $ written in terms of these two 
functions:\cite{Morii2004}
\begin{equation}
\label{transformation vacuum polarization function}
\begin{split}
\Pi_{QQ}(q^2)&=Q^2_1\Pi_{VV}(m^2_1,m^2_1,q^2)+Q^2_2\Pi_{VV}(m^2_2,m^2_2,q^2),\\
\Pi_{3Q}(q^2)&=\dfrac{1}{2}Q_1\Pi_{LV}(m^2_1,m^2_1,q^2)-\dfrac{1}{2}Q_2\Pi_{LV}(m^2_2,m^2_2,q^2),\\
&=\dfrac{1}{4}[Q_1\Pi_{VV}(m^2_1,m^2_1,q^2)-Q_2\Pi_{VV}(m^2_2,m^2_2,q^2)],\\
\Pi_{33}(q^2)&=\dfrac{1}{4}[\Pi_{LL}(m^2_1,m^2_1,q^2)+\Pi_{LL}(m^2_2,m^2_2,q^2)],\\
\Pi_{11}(q^2)&=\dfrac{1}{2}\Pi_{LL}(m^2_1,m^2_2,q^2).
\end{split}
\end{equation}

To proof \eqref{transformation vacuum polarization function} we use the definition of vacuum polarization functions and develop them in 
products of currents. For example, with $\Pi_{33}$ we have $J^{\mu}_3=\overline{L}\gamma_{\mu}\dfrac{\tau^3}{2}L$. Take product of 
$J^{\mu}_3$ and $J^{\mu}_3$ we have:

\begin{equation}
\begin{split}
 \Pi_{33}(q^2)&=\langle{J^{\mu}_{3},J^{\mu}_{3}}\rangle\\
 &=\dfrac{1}{4}[\langle{J^{\mu}_{Lt'},J^{\mu}_{Lt'}}\rangle+\langle{J^{\mu}_{Lb'},J^{\mu}_{Lb'}}\rangle]\\
 &=\dfrac{1}{4}[\Pi_{LL}(m^2_1,m^2_1,q^2)+\Pi_{LL}(m^2_2,m^2_2,q^2)],
 \end{split}
 \end{equation} 
with $J^{\mu}_{Lf}=\overline{f}\gamma^{\mu}(\dfrac{1-\gamma^5}{2}f)$. When $ \Pi_{LL} $ and $ \Pi_{LR} $ are summed up to get $ \Pi_{VV} $ 
and setting $ m_1=m_2=m $ we relevant the QED vacuum polarization function the quadratic term $ m^2 $ are canceled out and $\Pi_{VV}$ is 
proportional to $q^2$:
\begin{equation}
\Pi_{VV}(q^2)=q^2[-\dfrac{1}{2\pi^2}\int^1_0dxx(1-x)ln(\dfrac{\Lambda^2}{m^2-x(1-x)q^2})].
\end{equation}
This proportionality of $q^2$ follows from Ward identity. From the definition of $S$, $T$ and $U$ and \eqref{transformation} we obtain
\begin{equation}
\begin{split}
S &=\dfrac{1}{6{\pi}}[1-Yln(\dfrac{m^2_1}{m^2_2})],\\
T &=\dfrac{1}{16{\pi}s^2c^2m^2_Z}[m^2_1+m^2_2-\dfrac{2m^2_1m^2_2}{m^2_1-m^2_2}ln(\dfrac{m^2_1}{m^2_2})],\\
U &=\dfrac{1}{6\pi}[-\dfrac{5m^4_1-22m^2_1m^2_2+5m^4_2}{3(m^2_1-m^2_2)^2},\\
&-\dfrac{m^2_1-3m^4_1m^2_2-3m^2_1m^4_2+m^6_2}{(m^2_1-m^2_2)^3}ln(\dfrac{m^2_1}{m^2_2})].
\end{split}
\end{equation}
With case of the new generation of quarks, because they have the 3 colors we have: 
\begin{equation}
\begin{split}
S &=\dfrac{1}{2{\pi}}[1-Yln(\dfrac{m^2_1}{m^2_2})],\\
T &=\dfrac{3}{16{\pi}s^2c^2m^2_Z}[m^2_1+m^2_2-\dfrac{2m^2_1m^2_2}{m^2_1-m^2_2}ln(\dfrac{m^2_1}{m^2_2})],\\
U &=\dfrac{1}{2\pi}[-\dfrac{5m^4_1-22m^2_1m^2_2+5m^4_2}{3(m^2_1-m^2_2)^2},\\
&-\dfrac{m^2_1-3m^4_1m^2_2-3m^2_1m^4_2+m^6_2}{(m^2_1-m^2_2)^3}ln(\dfrac{m^2_1}{m^2_2})].
\end{split}
\end{equation}
From above expressions we see that $T$ is positive but $S$ and $U$ are not. However, if we assume that $\Delta{m}=|m_1-m_2|\ll m_1,m_2$, 
we obtain \cite{Peskin92}
\begin{equation}
\label{relation approximate}
\begin{split}
S & \thickapprox\dfrac{1}{6\pi},\\
T & \thickapprox\dfrac{1}{12{\pi}s^2c^2}[\dfrac{(\Delta{m})^2}{m^2_Z}],\\
U & \thickapprox\dfrac{2}{15\pi}[\dfrac{(\Delta{m})^2}{m^2_1}].
\end{split}
\end{equation}
This show that $S$ and $U$ are also positive in the limit where the isospin breaking in the doublet is small. Also as conjectured, $U$ is 
suppressed compared to $T$ by a factor of ($m^2_Z/m^2_1$). And because ${m_1,m_2}\gg{m_Z}$,contribution of new physics to $U$ parameters 
is very small. This is one reason that we usually do not consider the parameter $U$ when we search the effects of New Physics.

When we have one phenomenological model of physics beyond the SM with a new generation (e.g., $4^{th}$ generation), the fermions of this 
model will add contributions as the first and the second relation in \eqref{relation approximate} in $S$,  $T$ parameters. 
Therefore, $S$ can be used to measure of the total size of the new sector and $T$ is the measure of the total weak-isospin violating 
(difference of masses of two particles in one doublet) induced by it (because in \eqref{relation approximate} $T$ contains $\Delta{m}$). 
 
One another effect of T parameters is the effect of $m_t$, this effect is used to calculate the mass of top quark before we find it. 
Because $ m_t\gg m_b $ and $m_Z$ we have the contribution of $T$:
\begin{equation}
T \approx \dfrac{3}{64\pi^2\alpha}m^2_t,
\end{equation}
we can find the mass of the top quark.  

\chapter*{Conclusion}

During the period of doing this master thesis, I have studied the gauge theory, from the gauge invariance to spontaneous symmetry breaking 
and the Standard Model, which is presented in chapter 1. In chapter 2, I consdered loop corrections and calculated the vacuum polarization 
and when I calculated loop diagrams, I studied the Feynman parametrizions and regulization methods: the dimension regularization method and 
Pauli-Villars regularization method, two popular methods to loop calculations. In chapter 3, I studied the $S$,$T$,$U$ parameters which are  
very important for searching for the contributions of new physics, and check any model which predicts the new physics.

In this master thesis, I only presented contributions of $S$,$T$,$U$ parameters in the case of fermions of a new generation. In reality, to 
calculate $S$,$T$,$U$ we must calculate self-energies of photon, Z-boson, W-boson, and the mixing $\gamma-Z$ with all propagators. When we 
compute with all propagators, we will have contributions of Higgs bosons (dependending on the model there may be one Higgs- or many Higgs 
bosons) and vector fields.

This definition of $S$,$T$,$U$ parameters given above is only true if the gauge group of our model is $SU(2)_L \times U(1)_Y$, with other 
models which other gauge groups it needs modifications. 

Because the time of work is too short, the $S$,$T$,$U$ parameters in other models beyond SM (e.g., models with two Higgs doublets, Higgs 
triplet, right-handed neutrinos, etc.) will be investigated in the future. 

      
 \include{phuluc}
 
 
%\bibliographystyle{plain}
%\bibliography{tltk}

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\end{document}